Question: You have found the following ages (in years) of all 5 meerkats at your local zoo: $ 8,\enspace 2,\enspace 1,\enspace 2,\enspace 16$ What is the average age of the meerkats at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 meerkats at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{8 + 2 + 1 + 2 + 16}{{5}} = {5.8\text{ years old}} $ Find the squared deviations from the mean for each meerkat. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $8$ years $2.2$ years $4.84$ years $^2$ $2$ years $-3.8$ years $14.44$ years $^2$ $1$ year $-4.8$ years $23.04$ years $^2$ $2$ years $-3.8$ years $14.44$ years $^2$ $16$ years $10.2$ years $104.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{4.84} + {14.44} + {23.04} + {14.44} + {104.04}} {{5}} $ $ {\sigma^2} = \dfrac{{160.8}}{{5}} = {32.16\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{32.16\text{ years}^2}} = {5.7\text{ years}} $ The average meerkat at the zoo is 5.8 years old. There is a standard deviation of 5.7 years.